Comparing Economic Alternatives. Using time value of money to choose between two or more options. Examples of Engineering Decisions. Is it better to invest in an automated process or a manual process? Is it better to buy a high efficiency appliance or a low efficiency appliance?
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Using time value of money to choose
between two or more options
They differ because they have different cash flows.
Yet we must make a choice. Sometimes, we are allowed the choice of neither, but not always.
NPW costs =
1100 - 100 (P/F, 0.15, 10) + 200 (P/A, 0.15, 10)
NPW costs =
1300 - 100(P/F, 0.15, 10) + 150(P/A, 0.15, 10)
NPW > 0, so accept the extra investment
Choose the Challenger (Second) with the Smaller Net Annual Cost
Accept the extra investment of S over F. Choose S.
NPW = 35 + 35(P/A,0.15,3) = 114.91
NAC = 90(A/P,0.15,4)+ 10
NAC = 35(A/P,0.15,1)
= 40.25 per yr.
[-100(A/P, i, 9) + 20](P/A, i, 45) = 26.037
[-85(A/P, i, 5) + 24](P/A, i, 45) = 15.554
[-60(A/P, i, 5) + 18](P/A, i, 45) = 21.422
Select A since it is the alternative with the greatest NPW
-100(A/P, i, 9) + 20 = 2.64/year
-85(A/P, i, 5) + 24 = 1.57/year
-60(A/P, i, 5) + 18 = 2.17/year
1. Arrange (rank order) the feasible alternatives based on increasing initial investment.
2. Select the alternative with the smallest investment to be the defender (D).
3. From the remaining alternatives (not including the defender or those already eliminated) chose the challenger (C) as the alternative with the least initial investment.
4. Evaluate the increment of investment C – D by computing
NPW(C –D), NAW(C – D), ROR(C – D).
Accept of reject (C – D) based on the evaluation.
If (C – D) is accepted, delete D from the list, and make C the defender.
If (C – D) is rejected, delete C from the list.
If any alternatives remain, return to step 3, otherwise quit. The current defender is the best alternative.